Operaciones De Numeros Complejos

INSTITUTO POLITECNICO NACIONAL
ESIME
ICE

FUNDAMENTOS DEL ÁLGEBRA

ADOLFO NERI DEL RIO

Trabajo: Ejercicios DE NUMERO COMPLEJOS

Fecha: 7-septiembre-2011

2011

NUMEROS COMPLEJOSRESOLVER:

1i+21-i=1-i+2ii(1-i)=1+ii-i2=1+ii+1=1

2+i1-i+2-i1+i+i2=2+i1+i+2-i(1-i)1-i(1+i)+i2=2+2i+i+i2+2-2i-i+i21-i2+i2=22+i2= 1+i2=1+2i-1=2i

1+i+11+i2=1+i+11+i(1-i)(1-i)2= 1+i+1-i1+i-i-i22=2+2i+1-i22= 3+i22= 9+6i+i24= 8+6i4=2+3i2

¿CUAL ES EL MODULO DE –i+(3+i)/(1-i)?

-i+3+i1-i= -i+3+i1-i(1+i)(1+i)= -i+3+4i-11+1= -i+2+4i2= -i+2(1+2i)2=-i+1+2i= -i+1+2i=1+i

Z=12+12=2

DETERMINE3+4i5/(1+i3)

(3+4i)5(1+i3)=((3)2+(4)2)5(1)2+(3)2= (25)54=(5)25=1562.5


EVALUE (1+i)(3+i) Y CONVIERTA EL RESULTADO A LA FORMA CARTESIANA

(3+i+3i-1)= -1+3+i+3i= (-1+3)+i(1+3)= 0.73+i(2.73)EVALUE 1+i3+i-2-ii3+4i5+i EN LA FORMA CARTESIANA Y POLAR

1+i3+i-2-ii3+4i5+i= 3+4i-1-2-ii15+23i-4= 2+4i(-2-i)i(11+23i)= (-4-10i+4)11i-23= -10i11i-23(-11i-23)(-11i-23)=-110+230i121-253i+253i+529= -110+230i650= -0.16+0.35i carteciana
Z=(-016)2+(0.35)2=2613

Θ=tan-10.35-0.16+180=115.55°

Z=r(cos Θ+isen Θ)
=2613(cos 115.55° + i sen 115.55°)= -0.16+0.35i polares

ENCUENTRE LOSVALORES DE (-1)12= -1= i
(-1)12= -1= i

DETERMINE TODO LOS VALORES DE (3+i)15
Z=(3+i)15 5 Z5=3+i
|Z|=r= (3)2+(1)2=2 Θ=tan-113=16π n=5Zқ=r1ncosθ+2kπn+isenθ+2kπn;k=0,1,2,3,4

Z₀=r15cos16π+20π5+isen16π+20π5=1.14+0.12i

Z1=r15cos16π+21π5+isen16π+21π5=0.23+1.12i

Z2=r15cos16π+22π5+isen16π+22π5=-0.9+0.5i

Z3=r15cos16π+23π5+isen16π+23π5=-0.85-0.76iZ4=r15cos16π+24π5+isen16π+24π5=0.46-1.04i



(1+2i)19

|Z|= (1)2+(2)2=5 Θ=tan-121=63.43° n=19

(cos θ+isenθ)

(1+2i)19=519cis 63.43°=1953462.38+3906081.288iRESUELVE LAS SIG. EC.

a) w2-i=-1

w2=-1+i

|Z|= (-1)2+(1)2=2

Θ= tan-11-1=34π n=2

Zқ=r1ncosθ+2kπn+isenθ+2kπn;k=0,1

Z0=212cos 34π+20π2+isen...